1 M NH₄OH and 1 M HCI are mixed to make total volume of 300 mL. If pH of the mixture is 9.26 and pK_a(${\mathrm{NH}}_4^{+}$) = 9.26 then volume ratio of NH₄OH and HCl will be

pH = 9.26 indicates [NH₄OH] > [HCl] and thus, mixture is a buffer since, HCI will react with equivalent amount of NH₄OH forming NH₄Cl.

Let HCI = x mL x millimoles

$$\begin{gathered} {\mathrm{NH}}_4\mathrm{OH}=(300-\mathrm{x})\mathrm{mL}=(300-\mathrm{x})\text{ millimoles } \\ {\mathrm{NH}}_4\mathrm{Cl}\text{ formed }=\mathrm{x}\text{ millimoles } \\ {\mathrm{NH}}_4\mathrm{OH}\text{ unreacted }=300-\mathrm{x}-\mathrm{x}=(300-2\mathrm{x})\text{ millimoles } \\ \mathrm{pOH}=14-9.26=4.74 \\ {\mathrm{pK}}_{\mathrm{b}}=14-9.26=4.74 \\ \mathrm{pOH}={\mathrm{pK}}_{\mathrm{b}}+\log \frac{\left[{\mathrm{NH}}_4^{+}\right]}{\left[{\mathrm{NH}}_4\mathrm{OH}\right]} \\ 4.74=4.74+\log \frac{\mathrm{x}}{(300-2\mathrm{x})} \\ \frac{\mathrm{x}}{300-2\mathrm{x}}=1 \\ \therefore \mathrm{x}=100\mathrm{mL}=\text{ volume of }\mathrm{HCl} \\ (300-\mathrm{x})=200\mathrm{mL}=\text{ volume of }{\mathrm{NH}}_4\mathrm{OH}\text{ hence, }2:1 \end{gathered}$$